MaMaEuSch
Management Mathematics for
European Schools
http://www.mathematik.unikl.de/~mamaeusch/
Input-Output-Models and Markov-Chains
Ao. Univ.-Prof. Werner Peschek1
This project was published with the support of the EU by means of a partial support
within the Sokrates programm.
The content of this project does not necessarily reflect the standpoint of the EU neither is the
EU in any way responsible for the content.
1
Assistant at the didactics of mathematics department at the university of Klagenfurt
-1-
Input-Output-Models and Markov-Chains
I want to present two economic models, of which at least the first one is found in some
schoolbooks for commercial high schools (“HAK”) and general high schools (“AHS”).
I will start with an example for the:
Interconnection of Materials
Example:
A factory uses three components B1, B2, B3 for producing two intermediate
products Z1, Z2, from which in turn they produce two end products E1 and E2.
The required amounts for production are shown in the graph below:
 3 5


Interconnection matrix: (B→Z) =  8 4  ; (Z→E) =
 2 6


 7 8


 6 2
For instance, to produce 2 quantity units of Z1 and 3 quantity units of Z2
(demand NZ), you will need the following production demand XB of the
components B1, B2 and B3:
B1:
B2:
B3:
3⋅2 + 5⋅3 = 21
8⋅2 + 4⋅3 = 28
2⋅2 + 6⋅3 = 22
In matrix notation:
 21
 
XB = (B→
→Z) ⋅ NZ =  28 
 22 
 
XZ = (Z→
→E) ⋅ NE
corresponding:
-2-
 2
with NZ =  
 3
If you want to directly calculate a certain demand XB of components, for a
certain need NE of end products, you need the interconnection matrix (B→E).
As you can easily see from the graph, it is true that:
(B→E) = (B→Z) ⋅ (Z→E)
and thus
XB = (B→
→E) ⋅ NE = (B→
→Z) ⋅ (Z→
→E) ⋅ NE
In our example we get a need of E1 = 3 and E2 = 4
 3 5

  7 8   3
 ⋅   =
XB =  8 4  ⋅ 
6
2

  4
 2 6


 289 


 528 
 262 


This step by step treatment of the matter is getting a bit confusing if, apart from
the demand for end products, an additional demand for intermediate products
and/or components i.e. for a spare parts depot has to be satisfied and if this
fails, because there is a backflow in the interconnection of materials.
In such cases it’s better to present the whole interconnection of materials in one
interconnection matrix:
you need of
(in quantity units)
B1
B2
B3
Z1
Z2
E1
E2
For the production of a quantity unit of
B2
B3
Z1
Z2
B1
0
0
0
3
5
0
0
0
8
4
0
0
0
2
6
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
E1
0
0
0
7
6
0
0
E2
0
0
0
8
2
0
0
If you denote this interconnection matrix with A and the production demand of
B1, B2, ... , E2 with X, so
0

0
0

A⋅X =  0
0

0

0
0 0 3 5 0 0   x1 
  
0 0 8 4 0 0  x 2 
0 0 2 6 0 0  x3 
  
0 0 0 0 7 8 ⋅  x 4 
 
0 0 0 0 6 2   x 5 
0 0 0 0 0 0  x 6 
  
0 0 0 0 0 0  x 7 
indicates the internal (“secondary”) demand, resulting from the interconnection
of materials.
-3-
If you denote the external demand (“primary demand”) for B1, B2, ... , E2 with
N, so it’s true that:
X
=
N
+
A⋅X
Total demand (in quantity units) = primary demand (in quantity units) +
secondary demand (in quantity units)
In our example, for an external demand of 5 quantity units B1, 10 quantity units
B3, 5 quantity units Z1, 2 quantity units Z2, 3 quantity units E1 and 4 quantity
units E2 we get:
 x1   5   0 0 0
    
 x2   0  0 0 0
 x  10   0 0 0
 3   
 x4  =  5  + 0 0 0
x    
 5   2  0 0 0
 x6   3  0 0 0
    
 x7   4  0 0 0
3 5 0 0   x1 
  
8 4 0 0  x 2 
2 6 0 0  x3 
  
0 0 7 8 ⋅  x 4 
 
0 0 6 2   x 5 
0 0 0 0  x 6 
  
0 0 0 0  x 7 
This system of equations is easily and directly solved because A is a triangular
matrix. You can also find a general solving strategy by transforming the matrix
equation above:
X = N + A ⋅ X ⇔ X – A ⋅ X = N ⇔ (E – A) ⋅ X = N ⇔ X = (E – A)-1 ⋅ N
In our example:
1

0
0

E − A = 0
0

0

0
0 

0 
−2 −6 0
0 

1
0 − 7 − 8
0
1 − 6 − 2 
0
0
1
0 

0
0
0
1 
0 0 −3 −5
1 0 −8 −4
0
0
0
0
1
0
0
0
0 0
0
0
1

0
0

X = ( E − A) −1 ⋅ N =  0

0
0

0
(E − A ) −1
1

0
0

= 0
0

0

0
0 0 3 5 51 34 

1 0 8 4 80 72 
0 1 2 6 50 28 

0 0 1 0 7 8
0 0 0 1 6 2 
0 0 0 0 1 0

0 0 0 0 0 1
0 0 3 5 51 34   5   319 
   

1 0 8 4 80 72   0   576 
0 1 2 6 50 28  10   294 
   

0 0 1 0 7 8  ⋅  5  =  58 

   
0 0 0 1 6 2   2   28 
0 0 0 0 1 0 3  3 
   

0 0 0 0 0 1 4  4 
-4-
Open Leontief – Models
An economic system, i.e. a national economy, consists of several interdependent parts. They
are dependent on each other because for production each part needs services from their own
area as well as services from other parts (“secondary demand”) as “input”.
Moreover, there is usually also an external demand („primary demand“) of services from the
different parts.
As within the interconnection of materials, we are here dealing with the question, which
services (here measured in money units) each part has to perform, so that the primary and the
secondary demand are satisfied. Similar as for the interconnection of materials, it is true that:
total service (in money units) = primary demand (in money units) + secondary demand
(in money units)
Example:
A national economy may consist of three parts A, B, C.
From the following table you can see, which services (in money units) each
part needs, to be able to perform services itself, for a value of 1 money units:
for services with
the value of 1 money
units you need
A
B
C
from the parts
A
B
C
services with a value in (money units)
0,1
0,2
0,1
0,1
0,2
0,1
0
0,3
0,1
Within a certain period of time units we expect a demand of products from A,
B, C with a value of 1000, 300, 500 money units from all other economic
sectors (which do not belong to A, B or C).
Which services (in money units) have to be performed in A, B and C?
As with the interconnection of materials, it’s true that:
X = N + A ⋅ X that is X = (E – A)-1 ⋅ N
 0,1 0,1 0 


whereby A =  0,2 0,2 0,3 and N =
 0,1 0,1 0,1 


0 
 x A   0,9 − 0,1
  

thus  x B  =  − 0,2 0,8 − 0,3 
 x   − 0,1 − 0,1 0,9 
 C 

-5-
−1
1000


 300 
 500 


1000  1220 


 
⋅  300  =  980 
 500   800 

 

Linear models of the interconnection of materials and open Leontief-models have the same
structure:
There is an external demand (primary demand) for each product (services from each part), i.e.
altogether a demand vector N.
The interconnection of materials (the production dependencies) can be presented in one
matrix, A = (aij), whereby aij is the delivery (the output) of the product (the part) i for a
product (the part) j, that is, conversely the input from j out of i.
The amount (service) needed by each product (part) is represented by the vector X.
These models are called input-output-models and it’s true that:
X = N + A ⋅ X that is X = (E – A)-1 ⋅ N
The interconnection matrix A is thus also called input-output-matrix, the matrix (E – A)-1 is
called Leontief-matrix (Technology-matrix, Leontief-inverse).
Markov-Chains
Markov-Chains are results from random tests, whereby
-
the test result can only be one state (of many possible)
the test result is dependent on the initial state, as well as (known) probabilities for the
transition of one state to another
Example:
There is a competition of three providers (brands) A, B, C on the market.
From the following table you can see that from the buyers, who own a product
of the brand A today, 30% will buy A again within a certain period of time
units, 40% will change to brand B and 30% to brand C. The same
interpretation is for B and C.
from
A
B
C
A
0,3
0,4
0,1
to
B
0,4
0,1
0,3
C
0,3
0,5
0,6
We want to find out the share in the market that the brands A, B and C will
level off in the long run (if they will do so at all!), when the recent market
shares for A, B and C are 40%, 50% and 10%.
-6-
Initial situation: xA = 40%; xB = 50%;
xC = 10%
After one period of time the market shares are:
xA(new 1) = 0,3 ⋅ 0,4 + 0,4 ⋅ 0,5 + 0,1 ⋅ 0,1 = 33%
xB(new 1) = 0,4 ⋅ 0,4 + 0,1 ⋅ 0,5 + 0,3 ⋅ 0,1 = 24%
xC(new 1) = 0,3 ⋅ 0,4 + 0,5 ⋅ 0,5 + 0,6 ⋅ 0,1 = 43%
New initial situation: xA = 33%;
xB = 24%;
xC = 43%
After another period of time the market shares are:
xA(new 2) = 0,3 ⋅ 0,33 + 0,4 ⋅ 0,24 + 0,1 ⋅ 0,43 = 23,8%
xB(new 2) =
...........................................
= 28,5%
xC(new 2) = ..............................................
= 47,7%
Xt+1 = PT ⋅ Xt
whereby PT is the transpose of the matrix P =
(pij) and Xt resp. Xt+1 are the vectors of the
market shares at the points of time t resp. t+1.
In matrix notation:
From repeated iteration we assume that the market shares will tend towards a
balanced state:
xA ≈ 22,6%
xB ≈ 26,9%
xC ≈ 50,5%
That means, that for all products and each (further) period of time it’s true that:
market share old = market share new that means X = PT ⋅ X
I:
II:
III:
xA = 0,3xA + 0,4xB + 0,1xC
xB = 0,4xA + 0,1xB + 0,3xC
xC = 0,3xA + 0,5xB + 0,6xC
Obviously we have a homogeneous system of equations; we cancel the 3 parts,
but we add, that the sum of the market shares has to make 100%:
IV:
xA + xB + xC = 1
so that: 0,7xA – 0,4xB – 0,1xC = 0
-0,4xA + 0,9xB – 0,3xC = 0
xA + xB + xC = 1
xA ≈ 22, 58% ; xB ≈ 26,88% ; xC ≈ 50,54%
In our example, the achieved balanced state is independent from the choice of the initial
values – we are talking about ergodic Markov-chains.
-7-
This doesn’t always have to be the case. In the following, graphically depicted Markov-chain,
we also get a balanced state for every initial state, this however, is dependent on the choice of
the initial state – we are talking about non-ergodic Markov-chains.
Examples
1.
Given is the following graph („Gozinto-Graph“) of an interconnection of materials:
Develop a matrix A = (aij) which indicates, which amounts Ri are needed for the
production of a quantity unit of Zj (first production stage) and a matrix B = (bij), which
indicates, which amounts Zi are needed for the production of a quantity unit of Pj
(second production stage).
Calculate from A and B a matrix C = (cij) which indicates, which amounts Ri are
needed for the production of a quantity unit of Pj. Calculate which amounts of raw
materials R and which amounts of intermediate products Z are necessary to produce
200 quantity units of P1 and 400 quantity units of P2! In addition to the end products
P we need spare parts of 15, 10, 10 quantity units of Z1, Z2 resp. Z3. Calculate all
required amounts with the help of the Leontief-matrix!
2.
Given is the structure plan of quantities of a chemical
process. For A we have an external demand of 1000
quantity units, for B and C an external demand of 500
quantity units each.
Calculate the production amounts required in A-F!
3.
A mineral oil company is able to depose products with a value of 130.000 money units
daily. For the production of products with a value of one money unit they need coal
with a value of 0.1 money units, electricity with a value of 0.3 money units and diesel
oil (from their own production) with a value of 0.2 money units.
-8-
An electric power station has an external demand of 120.000 money units per day. To
perform services with a value of one money unit they require fuel with a value of 0.3
money units, coal with a value of 0.5 money units and power (from their own
production) with a value of 0.1 money units.
Finally, the coal mine, for producing coal with the value of one money unit, needs fuel
with the value of 0.1 money units and power with a value of 0.3 money units. The
daily, external demand is 20.000 money units.
With the help of the Leontief-inverse, calculate the daily amounts of production, so
that the internal and the external demand of these three companies can be satisfied!
4.
(Example by Bürger: “Mathematik Oberstufe 3”)
We divide the national economy of a country in
the following sectors, agriculture, trade and
commerce, industry and transportation.
The production and the use of goods (within a
certain period of time) are depicted in the figure
(in money units) aside.
We have the following external demand (in
money units) of goods from
agriculture:
18
trade and commerce: 12
industry:
8
transportation:
2
Transportation
With the help of the Leontief-inverse, calculate the amount, that has to be produced in
each sector! Interpret the results!
5.
Calculate the market shares to which the companies
R, S and T will level off in the long run, when the
consumer behaviour corresponds to a Markov-chain
with the given probabilities of changeover.
Follow the development of the market shares for
the Markov-chain given in the graph, through
three periods of time, where we start from
a) 70%, 10%, 10%, 10%
b) 10%, 10%, 10%, 70%
market shares for A, B, C resp. D!
Try to forecast a balanced state, independent from the initial state! What do you realize
from it?
-9-